1.

How many odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed?

Answer»

Given : 

Odd number less than 1000 is required. 

In order to make the number odd, the last digit has to either of (3, 5, 7)

(No Zero)(Any 3 digits left(3,5,or 7)

Assume three boxes,

In the first either of the three digits (3,5,7) can be placed, so the possibility is 3C1 

Case 1: Middle digit is zero 

If the middle digit is zero, number of ways of placing odd numbers on the second box = 2 

Hence, 

The total number of ways = 3 × 2 = 6 ways 

Case 2 : Middle digit is an odd number Number of ways of filling middle box = 2 

Number of ways of filling third box = 1 

Hence, 

The total number of ways = 3 × 3 

= 9 ways 

Hence,

Total number of outcomes possible = 6 + 9 

= 15 ways



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